Answer:
D) ¼ as large
Explanation:
The magnitude of the electric force between two charges is
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the magnitudes of the two charges
r is the separation between them
From the formula we see that the magnitude of the force is inversely proportional to the square of the distance.
In this problem, the distance is doubled:
[tex]r'=2r[/tex]
So the new force will be
[tex]F'=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}k\frac{q_1 q_2}{r^2}=\frac{1}{4}F[/tex]
So the force will decrease to 1/4 of its original value.
